Question

The equation of the line which is parallel to the lines common to the pair of lines given by $6{x^2} - xy - 12{y^2} = 0$ and $15{x^2} + 14xy - 8{y^2} = 0$ and 5 units away from the origin and having positive $x$ and $y$ intercepts is:
A) $4x + 3y = 12$
B) $3x + 4y = 25$
C) $12x + 5y = 65$
D) $15x + 8y = 85$

Answer 1

We will find the line that is common to $6{x^2} - xy - 12{y^2} = 0$ and $15{x^2} + 14xy - 8{y^2} = 0$. We will find the line’s slope. The slope of the required line will also be the same. We will assume that the line has constant $k$. We will find the line’s distance from the origin and equate it with 5. We will find the value of $k$ using this equation. We will check if the line has positive intercepts.

Formulas used:
1. Quadratic equation $a{x^2} + b + c = 0$ has the roots $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
2. Distance of a line $ax + by + c = 0$ from a point $\left( {x,y} \right)$ is given by $\dfrac{{\left| {ax + by + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$.
3. Slope of a line $ax + by + c = 0$ is given by $m = \dfrac{{ - a}}{b}$.
4. The intercept form of a line is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$ where $a$ and $b$ are the $x$ and $y$ intercepts respectively.

Complete step by step solution:
We will find the lines represented by the pair of lines $6{x^2} - xy - 12{y^2} = 0$. We will substitute 6 for $a$, $- y$ for $b$ and $- 12{y^2}$ for $c$ in the first formula:
$x = \dfrac{{ - \left( { - y} \right) \pm \sqrt {{{\left( { - y} \right)}^2} - 4 \cdot 6\left( { - 12{y^2}} \right)} }}{{2 \cdot 6}}$
We will simplify the above equation:
$\begin{array}{l}x = \dfrac{{y \pm \sqrt {{y^2} + 288{y^2}} }}{{12}}\\\ \Rightarrow x = \dfrac{{y \pm 17y}}{{12}}\\\ \Rightarrow x = \dfrac{{18y}}{{12}},\dfrac{{ - 16y}}{{12}}\\\ \Rightarrow x = \dfrac{{3y}}{2}, - \dfrac{{4y}}{3}\end{array}$
$\begin{array}{l}x = \dfrac{{3y}}{2}\\\ \Rightarrow 2x = 3y\\\ \Rightarrow 2x - 3y = 0\end{array}$
$\begin{array}{l}x = - \dfrac{{4y}}{3}\\\ \Rightarrow 3x = - 4y\\\ \Rightarrow 3x + 4y = 0\end{array}$
$\therefore$ The pair of lines $6{x^2} - xy - 12{y^2} = 0$ represents the line $2x - 3y = 0$ and the line $3x + 4y = 0$.
We will find the lines represented by the pair of lines $15{x^2} + 14xy - 8{y^2} = 0$. We will substitute 15 for $a$, $14y$ for $b$ and $- 8{y^2}$ for $c$ in the first formula:
$x = \dfrac{{ - \left( {14y} \right) \pm \sqrt {{{\left( { - 14y} \right)}^2} - 4 \cdot 15\left( { - 8{y^2}} \right)} }}{{2 \cdot 15}}$
We will simplify the above equation:
$\begin{array}{l}x = \dfrac{{ - 14y \pm \sqrt {196{y^2} + 480{y^2}} }}{{30}}\\\ \Rightarrow x = \dfrac{{ - 14y \pm 26y}}{{30}}\\\ \Rightarrow x = \dfrac{{ - 14y + 26y}}{{30}},\dfrac{{ - 14y - 26y}}{{30}}\\\ \Rightarrow x = \dfrac{{12y}}{{30}},\dfrac{{ - 40y}}{{30}}\end{array}$
$\begin{array}{l}x = \dfrac{{12y}}{{30}}\\\ \Rightarrow x = \dfrac{{2y}}{5}\\\ \Rightarrow 5x = 2y\\\ \Rightarrow 5x - 2y = 0\end{array}$
$\begin{array}{l}x = - \dfrac{{40y}}{{30}}\\\ \Rightarrow x = - \dfrac{{4y}}{3}\\\ \Rightarrow 3x = - 4y\\\ \Rightarrow 3x + 4y = 0\end{array}$
$\therefore$ The pair of lines $15{x^2} + 14xy - 8{y^2} = 0$ represents the line $5x - 2y = 0$ and the line $3x + 4y = 0$.
We can see that the line common to both the pairs is $3x + 4y = 0$.
We will find the slope of the line. We will substitute 3 for $a$ and 4 for $b$ in the third formula:
$m = - \dfrac{3}{4}$
We know that the slope of the required line will also be $- \dfrac{3}{4}$.
We will assume that the line is $ax + by + c = 0$. As the slope of the line is $- \dfrac{3}{4}$, we can assume that the equation of the line will be $3x + 4y + k = 0$.
We will find the distance of origin from this line and equate it with 5. We will substitute 0 for $x$, $y$; $- 3$ for $a$ and 4 for $b$ in the 2nd formula:

\Rightarrow k = 25, - 25\end{array}$$ The 2 possibilities for the equation of line are: $$\begin{array}{l}{\rm{ }}3x + 4y + 25 = 0\\\ \Rightarrow \Rightarrow {\rm{ }} - 3x - 4y = 25\end{array}$$ and $$\begin{array}{l}3x + 4y - 25 = 0\\\ \Rightarrow \Rightarrow {\rm{ }}3x + 4y = 25\end{array}$$ We will divide the equations of both the lines by 25 to convert them into the intercept form. $$\begin{array}{l}{\rm{ }} - \dfrac{{3x}}{{25}} - \dfrac{{4x}}{{25}} = \dfrac{{25}}{{25}}\\\ \Rightarrow \dfrac{x}{{ - 3/25}} + \dfrac{y}{{ - 4/25}} = 1\end{array}$$ $$\begin{array}{l}{\rm{ }}\dfrac{{3x}}{{25}} + \dfrac{{4y}}{{25}} = \dfrac{{25}}{{25}}\\\ \Rightarrow \dfrac{x}{{3/25}} + \dfrac{y}{{4/25}} = 1\end{array}$$ We can see by comparing with the fourth formula that the 1st equation has negative intercepts while the 2nd equation has positive intercepts. $$\therefore 3x + 4y = 25$$ is the equation of the required line. **Option B is the correct option.** **Note:** We can also find the solution by checking which equation from the option satisfies the conditions given in the question. The equation given in option A does not have a distance of 5 units from the origin. So, option A is eliminated. Out of options B, C and D; the equations in option C and D do not satisfy the pair of lines given in the question. So, they are also eliminated. We can conclude that option B is the correct option.